Question

find all second-order partial derivatives at (0,0) of the function f(x,y)=
x2tan−1yx−y2tan−1xy,x≠0,y≠0

Answer 1

Can you name what the second-order partial derivatives are I'll give an example of one

\[F_{xx}\]

Answer 2

\[f(x)=x^2tan(y)-1yx-y^2tanx-1xy=x^2tan(y)-2xy-y^2tanx\] is this your equation you're missing some stuff

Answer 3

such as the variables for your tangent functions

Answer 4

its x squared tan inverse y over x minus y squared tan inverse x over y

Answer 5

\[x^2tan^{-1}(\frac{y}{x})-y^2tan^{-1}(\frac{x}{y})\]

the first you'll have to use a product rule

\[uv'+vu'=2xtan^{-1}(\frac{y}{x})+x^2\frac{d}{dx}tan(\frac{y}{x})\]

Answer 6

then you have to think of y as a constant for tan inverse

Answer 7

\[\frac{d}{dx}tan^{-1}\frac{c}{x}\]

Answer 8

can you refresh me of what tan^{-1} derivative is



\[\frac{u'}{u^2+x^2}\]?

Answer 9

\[\frac{ d }{ dx }\tan^{-1} x=\frac{ 1 }{ 1+x^{2} }\]

Answer 10

Evaluate the iterated integral \[\int\limits_{0}^{a}xdx \int\limits_{0}^{\sqrt{a^{2}-x ^{2}}}\int\limits_{0}^{\sqrt{a ^{2}-x ^{2}-y ^{2}}}zdz,a>0\]

Answer 11

so your

\[u=c(\frac{1}{x})\]
\[u'=-cx^{-2}=\frac{-c}{x^2}\]

\[\frac{d}{dx}(tan^{-1}\frac{y}{x})=\frac{1}{(\frac{y}{x})^2+x^2}*\frac{-y}{x^2}\]